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How many three-digit numbers ABC, in which A, B, and C are each digits

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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 08 Jul 2016, 20:44
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How many three-digit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?

A. 33
B. 36
C. 41
D. 45
E. 50
[Reveal] Spoiler: OA

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Senior Manager
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Re: How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 09 Jul 2016, 00:25
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first 3 digit number which qualifies for 2B=A+C > 111
next is 123
next is 135
gap of 12
900 3 digit numbers
so 900/12=50

E?

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Re: How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 09 Jul 2016, 02:12
Possible solutions are
Diff of 0 --
111,222,........,999 = 9
Diff of 1 --
123, 234, ....., 789 =7x2 = 14
Diff of 2 --
135, 246.......579 = 5x2 = 10
Diff of 3 --
147,258,369 = 3x2 = 6
Diff of 4 --
159 = 1x2 = 2
Last
210, 420,630,840 = 4

So total is 9+14+10+6+2+4 = 45.

Not a good way to solve.
Experts please

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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 09 Jul 2016, 04:34
2B = A+C
Middle number can only be 2, 4,6,8,10,12,14,16,18
now when middle number is 2 , you can have only 1 combination A =1 and C =1

when middle is 4 you can have two pairs 1 ,3 and 2,2
when middle is 6 you can have three
so basically 1+2+3+4+5+6+7+8+9 = 45
D is the answer

I ma confused about one thing , why we are not required to consider op pair
when we have 4 at middle we can have (1,3 ) (2,2)and (3,1)
when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51
another series is 1+3+5+...17 == 81 , which is not answer

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Re: How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 10 Jul 2016, 01:21
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2B = A + C
Let A=C
therefore 2A = 2B
therefore A=B=C is part of solution.
There are 9 such occurrences

now let A = 1, possible values of c =3,5,7,9
A = 2, possible values of c =4,6,7,0
A = 3, possible values of c =1,5,7,9
A = 4, possible values of c =2,6,8,0
... so on...
There are 9 such rows = 9 * 4 = 36
Total = 36 + 9 = 45
D

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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 07 Oct 2016, 21:20
I just did manual combinations.. Though I got it right, it took lot of time..
A-B-C
Start with number 9 for B..

A-9-C --> we have to find combinations of A and C such that A+C=18 ----> this yields only one combination --> 9-9-9
A-8-C --> we have to find combinations of A and C such that A+C=16 ----> this yields three combinations --> 8-8-8, 9-8-7, 7-9-9
..
..
..
..

Do the same for all digits and make the list
B---- num of combinations
9---- 1
8---- 3
7---- 5
6---- 7
5---- 9
4---- 8
3---- 6
2---- 4
1---- 2
0---- 0
-------------
Total ---- 45 (add all above numbers in red color)

Answer choice D is correct..
But this procedure takes lot of time...

Is there any quick solution?

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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 28 Oct 2016, 01:33
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2B=A+C is the same as B=(A+C)/2. That is B is the arithmetic mean of the other 2 digits.
This can be achieved only in case, when parity of both A and C is the same. So we have odd + odd or even + even.
Let's look at first case. We have 5 ways to choose odd number for A and same 5 ways to choose odd number for C - total 5*5=25.
In the case of even numbers, we can't choose 0 for A (otherwise it won't be 3 digit number) so we have 4*5=20. Those cases are independent so in total we have 45 ways.
Answer is D.

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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 28 Oct 2016, 14:53
alpham wrote:
How many three-digit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?

A. 33
B. 36
C. 41
D. 45
E. 50


If we assume numbers that work are distributed
equally through the 9 100s blocks,
only answers B and D are multiples of 9.
If we choose any 100s block, say the 500s, and start at 555,
which we know will work, and count forward and backward
by intervals of 12 until numbers no longer satisfy,
we will have 5 numbers that work: 555,567,579,543,531.
5*9=45
D

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Re: How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 13 Nov 2017, 04:42
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vitaliyGMAT wrote:
2B=A+C is the same as B=(A+C)/2. That is B is the arithmetic mean of the other 2 digits.
This can be achieved only in case, when parity of both A and C is the same. So we have odd + odd or even + even.
Let's look at first case. We have 5 ways to choose odd number for A and same 5 ways to choose odd number for C - total 5*5=25.
In the case of even numbers, we can't choose 0 for A (otherwise it won't be 3 digit number) so we have 4*5=20. Those cases are independent so in total we have 45 ways.
Answer is D.



Alternatively, we are given that
2B = A + C
So whatever A and C are, they add up to give an even sum. Hence either A and C are both even or both odd.
Now proceed as given above.
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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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alpham wrote:
How many three-digit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?

A. 33
B. 36
C. 41
D. 45
E. 50

1.Interpret what is given so that we get basic info. 2B = A + C implies that A and C are both odd or both even
2. When A and C are both odd, A can be 1,3,5,7,9 and for each of these B can be 1,3,5,7,9 for a total of 25 cases
3. When A and C are both even A can be 2,4,6,8 and for each of these B can be 0,2,4,6,8 for a total of 20 cases
4. So the answer is 25+20= 45 cases
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Re: How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 16 Nov 2017, 12:32
mbaprep2016 wrote:
2B = A+C
Middle number can only be 2, 4,6,8,10,12,14,16,18
now when middle number is 2 , you can have only 1 combination A =1 and C =1

when middle is 4 you can have two pairs 1 ,3 and 2,2
when middle is 6 you can have three
so basically 1+2+3+4+5+6+7+8+9 = 45
D is the answer

I ma confused about one thing , why we are not required to consider op pair
when we have 4 at middle we can have (1,3 ) (2,2)and (3,1)
when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51
another series is 1+3+5+...17 == 81 , which is not answer


I also have the same doubt. Can some expert help here?
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How many three-digit numbers ABC, in which A, B, and C are each digits [#permalink]

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New post 16 Nov 2017, 15:33
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GMATisLovE wrote:
mbaprep2016 wrote:
2B = A+C
Middle number can only be 2, 4,6,8,10,12,14,16,18
now when middle number is 2 , you can have only 1 combination A =1 and C =1

when middle is 4 you can have two pairs 1 ,3 and 2,2
when middle is 6 you can have three
so basically 1+2+3+4+5+6+7+8+9 = 45
D is the answer

I ma confused about one thing , why we are not required to consider op pair
when we have 4 at middle we can have (1,3 ) (2,2)and (3,1)
when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51
another series is 1+3+5+...17 == 81 , which is not answer


I also have the same doubt. Can some expert help here?


Hi,

We have 2B=A+C as follows:

when middle number is 1 we have the pairs (1,1) (2,0)
When middle number is 2 we have the pairs (2,2 ) (3,1) (1,3 ) (4,0)
When middle number is 3 we have the pairs (3,3) (4,2), (2,4) (5,1) (1,5) (6,0)
When middle number is 4 we have the pairs (4,4 ) (5,3) (3,5) (6,2) (2,6) (1,7) (7,1) (8,0)
When middle number is 5 we have the pairs (5,5) (6,4,) (4,6) (7,3) (3,7) (8,2) (2,8) (9,1) (1,9)
When middle number is 6 we have the pairs (6,6 ) (7,5) (5,7) (4,8 ) (8,4 ) (9,3) (3,9 )
When middle number is 7 we have the pairs (7,7) (8,6) (6,8) (9,5) (5,9)
When middle number is 8 we have the pairs (8,8 ) (9,7) (7,9)
When middle number is 9 we have the pairs ( 9,9)

for a total of 45 cases
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How many three-digit numbers ABC, in which A, B, and C are each digits   [#permalink] 16 Nov 2017, 15:33
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How many three-digit numbers ABC, in which A, B, and C are each digits

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